Question

If the solution of the system of simultaneous equations \[ \frac{1}{x} + \frac{2}{y} - \frac{3}{z} = 1, \quad \frac{2}{x} - \frac{4}{y} + \frac{3}{z} = 1, \quad \frac{3}{x} + \frac{6}{y} - \frac{6}{z} = 4 \] is \( x = \alpha, y = \beta, z = \gamma \), then the value of \( \alpha^2 + \gamma^2 \) is:

• A. \(5\beta\)
• B. \(\beta^2\)
• C. \(3\beta\)
• D. \(2\beta^2\)

Hint:

For nonlinear equations involving reciprocals, use variable substitution (e.g., \( u = \frac{1}{x} \)) to convert into a linear system and solve systematically.

Answer 1

The Correct Option is A

Let us define: \[ u = \frac{1}{x}, \quad v = \frac{1}{y}, \quad w = \frac{1}{z} \] Rewriting the system: \begin{align} u + 2v - 3w = 1 \quad \text{(1)}
2u - 4v + 3w = 1 \quad \text{(2)}
3u + 6v - 6w = 4 \quad \text{(3)} \] Now solve the system of equations. Multiply (1) by 2: \[ 2u + 4v - 6w = 2 \quad \text{(4)} \] Add (4) and (2): \[ (2u - 4v + 3w) + (2u + 4v - 6w) = 1 + 2 \Rightarrow 4u - 3w = 3 \quad \text{(5)} \] Multiply (1) by 3: \[ 3u + 6v - 9w = 3 \quad \text{(6)} \] Subtract (3): \[ (3u + 6v - 9w) - (3u + 6v - 6w) = 3 - 4 \Rightarrow -3w = -1 \Rightarrow w = \frac{1}{3} \Rightarrow z = 3 \] Now use (1): \[ u + 2v - 1 = 1 \Rightarrow u + 2v = 2 \quad \text{(7)} \] Use (2) and known \( w = \frac{1}{3} \Rightarrow 3w = 1 \): \[ 2u - 4v + 1 = 1 \Rightarrow 2u - 4v = 0 \Rightarrow u = 2v \quad \text{(8)} \] Substitute (8) in (7): \[ 2v + 2v = 2 \Rightarrow 4v = 2 \Rightarrow v = \frac{1}{2} \Rightarrow y = 2,\quad u = 1 \Rightarrow x = 1 \] Now, \( \alpha = 1, \beta = 2, \gamma = 3 \Rightarrow \alpha^2 + \gamma^2 = 1^2 + 3^2 = 10 = 5\beta \)